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Given a matrix $M\in A^{n\times n}$, where $A$ is a commutative ring different from $\{0\}$, then we know that if there exists a vector $x\in A^n$ such that $Mx=0$, then $\det M$ must be a zero divisor.

The converse is true? That is, if $\det M$ is a divisor of zero, is it true that his kernel is non-trivial? And if $\det M=0$?

The only thing I obtained is that it's true for Principal Ideal Domains, thanks to the Smith Form.

Ben Grossmann
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Exodd
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1 Answers1

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The below is incorrect as of now, but might help someone give a correct proof. Feel free to modify it.

If $\det(M)$ is a zero divisor, we have the following: let $a \in A$ be such that $a\det(M) = 0$, and take $x = (a,0,\dots,0)^T$. We have $$ \DeclareMathOperator{\adj}{adj} M[\adj(M)x] = [\det(M)\adj(M)] x = \det(M) Ix = 0 $$ Thus, if $\det(M)$ is a zero divisor, the kernel of $M$ is indeed necessarily non-trivial.

Ben Grossmann
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  • not quite: who tells you that $adj(M)x\ne 0$? – Exodd May 29 '15 at 21:59
  • You should prove that $\operatorname{adj}(M)x\ne0$. – egreg May 29 '15 at 22:00
  • We know that $\det (M)I$ is non-zero, so $\adj(M)$ can't be zero. So, if it's $j$th column is non-zero, put $a$ in the $j$th entry. – Ben Grossmann May 29 '15 at 22:06
  • Let $M=\left[\begin{smallmatrix}a_{11}&a_{12}\a_{21}&a_{22}\end{smallmatrix}\right]$ be such that $Mb=0$ for some $b\ne0$ and that $\det(M)\ne0$; note that $\det(M)b=0$. Then for $x=\left[\begin{smallmatrix}b\0\end{smallmatrix}\right]$ you have $\operatorname{adj}(M)x=0$. – egreg May 29 '15 at 22:08
  • @egreg perhaps that complicates things... – Ben Grossmann May 29 '15 at 22:37